Prove that $int_0^x fracsin tt dt > arctan x $ for $x>0$.
Clash Royale CLAN TAG#URR8PPP
up vote
8
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I'm finding some bounds for the Si function defined as
$$
operatornameSi(x) := int_0^xfracsin ttdt.
$$
I observed from WolframAlpha that the inequality
$$
operatornameSi(x)>arctan(x)
$$
holds for $x>0$.
I tried to show this analytically but failed and could not find any references regarding this. Could someone help me with this?
calculus inequality
asked Aug 11 at 3:20
Ramanasa
564
 |Â
show 10 more comments
up vote
8
down vote
favorite
I'm finding some bounds for the Si function defined as
$$
operatornameSi(x) := int_0^xfracsin ttdt.
$$
I observed from WolframAlpha that the inequality
$$
operatornameSi(x)>arctan(x)
$$
holds for $x>0$.
I tried to show this analytically but failed and could not find any references regarding this. Could someone help me with this?
calculus inequality
asked Aug 11 at 3:20
Ramanasa
564
I don't know if this helps but maybe try to express $arctan(x)$ as the integral from $0$ to $x$ of its derivative $frac1t^2+1$ and then form one integral.
– zzuussee
Aug 11 at 3:26
I tried to but could not proceed more. It seems there needs some more manipulations.
– Ramanasa
Aug 11 at 3:34
I'm currently writing a sketch of some thoughts.
– zzuussee
Aug 11 at 3:34
If you can deal with power series, then write both $operatornameSi(x)$ and $arctan (x)$ as power series and note their difference when $x>0.$
– Allawonder
Aug 11 at 3:43
2
@zzuussee: but a faulty one, since $arctan(x)$ is not an entire function. On the other hand an interesting relation between $arctan$ and $textSi$ is $$ mathcalLleft(textSi(x)right)(s) = frac1sarctanfrac1s.$$
– Jack D'Aurizio♦
Aug 11 at 3:52
 |Â
show 10 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I'm finding some bounds for the Si function defined as
$$
operatornameSi(x) := int_0^xfracsin ttdt.
$$
I observed from WolframAlpha that the inequality
$$
operatornameSi(x)>arctan(x)
$$
holds for $x>0$.
I tried to show this analytically but failed and could not find any references regarding this. Could someone help me with this?
calculus inequality
asked Aug 11 at 3:20
Ramanasa
564
I'm finding some bounds for the Si function defined as
$$
operatornameSi(x) := int_0^xfracsin ttdt.
$$
I observed from WolframAlpha that the inequality
$$
operatornameSi(x)>arctan(x)
$$
holds for $x>0$.
I tried to show this analytically but failed and could not find any references regarding this. Could someone help me with this?
calculus inequality
asked Aug 11 at 3:20
Ramanasa
564
asked Aug 11 at 3:20
Ramanasa
564
asked Aug 11 at 3:20
Ramanasa
564
asked Aug 11 at 3:20
Ramanasa
564
564
I don't know if this helps but maybe try to express $arctan(x)$ as the integral from $0$ to $x$ of its derivative $frac1t^2+1$ and then form one integral.
– zzuussee
Aug 11 at 3:26
I tried to but could not proceed more. It seems there needs some more manipulations.
– Ramanasa
Aug 11 at 3:34
I'm currently writing a sketch of some thoughts.
– zzuussee
Aug 11 at 3:34
If you can deal with power series, then write both $operatornameSi(x)$ and $arctan (x)$ as power series and note their difference when $x>0.$
– Allawonder
Aug 11 at 3:43
2
@zzuussee: but a faulty one, since $arctan(x)$ is not an entire function. On the other hand an interesting relation between $arctan$ and $textSi$ is $$ mathcalLleft(textSi(x)right)(s) = frac1sarctanfrac1s.$$
– Jack D'Aurizio♦
Aug 11 at 3:52
 |Â
show 10 more comments
I don't know if this helps but maybe try to express $arctan(x)$ as the integral from $0$ to $x$ of its derivative $frac1t^2+1$ and then form one integral.
– zzuussee
Aug 11 at 3:26
I tried to but could not proceed more. It seems there needs some more manipulations.
– Ramanasa
Aug 11 at 3:34
I'm currently writing a sketch of some thoughts.
– zzuussee
Aug 11 at 3:34
If you can deal with power series, then write both $operatornameSi(x)$ and $arctan (x)$ as power series and note their difference when $x>0.$
– Allawonder
Aug 11 at 3:43
2
@zzuussee: but a faulty one, since $arctan(x)$ is not an entire function. On the other hand an interesting relation between $arctan$ and $textSi$ is $$ mathcalLleft(textSi(x)right)(s) = frac1sarctanfrac1s.$$
– Jack D'Aurizio♦
Aug 11 at 3:52
I don't know if this helps but maybe try to express $arctan(x)$ as the integral from $0$ to $x$ of its derivative $frac1t^2+1$ and then form one integral.
– zzuussee
Aug 11 at 3:26
I don't know if this helps but maybe try to express $arctan(x)$ as the integral from $0$ to $x$ of its derivative $frac1t^2+1$ and then form one integral.
– zzuussee
Aug 11 at 3:26
I tried to but could not proceed more. It seems there needs some more manipulations.
– Ramanasa
Aug 11 at 3:34
I tried to but could not proceed more. It seems there needs some more manipulations.
– Ramanasa
Aug 11 at 3:34
I'm currently writing a sketch of some thoughts.
– zzuussee
Aug 11 at 3:34
I'm currently writing a sketch of some thoughts.
– zzuussee
Aug 11 at 3:34
If you can deal with power series, then write both $operatornameSi(x)$ and $arctan (x)$ as power series and note their difference when $x>0.$
– Allawonder
Aug 11 at 3:43
If you can deal with power series, then write both $operatornameSi(x)$ and $arctan (x)$ as power series and note their difference when $x>0.$
– Allawonder
Aug 11 at 3:43
2
2
@zzuussee: but a faulty one, since $arctan(x)$ is not an entire function. On the other hand an interesting relation between $arctan$ and $textSi$ is $$ mathcalLleft(textSi(x)right)(s) = frac1sarctanfrac1s.$$
– Jack D'Aurizio♦
Aug 11 at 3:52
@zzuussee: but a faulty one, since $arctan(x)$ is not an entire function. On the other hand an interesting relation between $arctan$ and $textSi$ is $$ mathcalLleft(textSi(x)right)(s) = frac1sarctanfrac1s.$$
– Jack D'Aurizio♦
Aug 11 at 3:52
 |Â
show 10 more comments
2 Answers
2
active
oldest
votes
up vote
6
down vote
All right, I realized that representing $arctan(x)$ through the integral of an oscillating function is not a good idea. Better to represent both $arctan(x)$ and $textSi(x)$ as integrals of monotonic and easily-comparable functions. So here it is a polished version of the previous answer. We may safely assume $x>1$ since power series easily prove the statement for $xin[0,1]$. By the Laplace transform and the Cauchy-Schwarz inequality
$$ textSi(x)=fracpi2-int_0^+inftyfraccos(x)+ssin(x)(1+s^2)e^sx,dsgeq fracpi2-int_0^+inftyfracdse^sxsqrt1+s^2. tag1$$
By the very definition of $arctan$ we have $arctan(x)=left(int_0^+inftyfrac11+s^2-frac11+(s+x)^2right),dx$, hence through $arctan x=fracpi2-arctanfrac1x$ we get the following integral representation:
$$ arctan(x) = fracpi2-int_0^+inftyfrac1+2sx(1+s^2)(1+2sx+x^2+s^2 x^2),ds. tag2$$
For the sake of brevity, let us denote as $S(x,s)$ and $T(x,s)$ the integrand functions appearing in the RHSs of $(1)$ and $(2)$. If we manage to prove $S(x,s)leq T(x,s)$ for any $x>1$ and any $s>0$ we are done. But the Padé approximants for the exponential function reveal that this is a pretty loose inequality, so we are good to go:
$$ forall x>0,qquad textSi(x)>arctan(x).tag3 $$
A strange-looking consequence of $(1)$ and the AM-QM inequality is also
$$ textSi(x) > fracpi2-sqrt2,e^x,Gamma(0,x).tag4$$
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
 |Â
show 8 more comments
up vote
0
down vote
$$textSi'(x) = fracsin xx = fracsum_k (-1)^kx^2k+1/(2k+1)!x = sum_k (-1)^kx^2k/(2k+1)!$$
$$arctan'(x) = frac11+x^2 = frac11-(-x^2) = sum_k (-x^2)^k = sum_k (-1)^kx^2k$$
$$textSi(x) = sum_k frac(-1)^kx^2k+1/(2k+1)(2k+1)!$$
$$arctan(x) = sum_k frac(-1)^kx^2k+1/(2k+1)1$$
$$frac1(2k+1)! geq frac11$$
This would give the opposite of your inequality... but we must take into account the $(-1)^k$.
answered Aug 11 at 3:42
mr_e_man
854420
3
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
All right, I realized that representing $arctan(x)$ through the integral of an oscillating function is not a good idea. Better to represent both $arctan(x)$ and $textSi(x)$ as integrals of monotonic and easily-comparable functions. So here it is a polished version of the previous answer. We may safely assume $x>1$ since power series easily prove the statement for $xin[0,1]$. By the Laplace transform and the Cauchy-Schwarz inequality
$$ textSi(x)=fracpi2-int_0^+inftyfraccos(x)+ssin(x)(1+s^2)e^sx,dsgeq fracpi2-int_0^+inftyfracdse^sxsqrt1+s^2. tag1$$
By the very definition of $arctan$ we have $arctan(x)=left(int_0^+inftyfrac11+s^2-frac11+(s+x)^2right),dx$, hence through $arctan x=fracpi2-arctanfrac1x$ we get the following integral representation:
$$ arctan(x) = fracpi2-int_0^+inftyfrac1+2sx(1+s^2)(1+2sx+x^2+s^2 x^2),ds. tag2$$
For the sake of brevity, let us denote as $S(x,s)$ and $T(x,s)$ the integrand functions appearing in the RHSs of $(1)$ and $(2)$. If we manage to prove $S(x,s)leq T(x,s)$ for any $x>1$ and any $s>0$ we are done. But the Padé approximants for the exponential function reveal that this is a pretty loose inequality, so we are good to go:
$$ forall x>0,qquad textSi(x)>arctan(x).tag3 $$
A strange-looking consequence of $(1)$ and the AM-QM inequality is also
$$ textSi(x) > fracpi2-sqrt2,e^x,Gamma(0,x).tag4$$
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
 |Â
show 8 more comments
up vote
6
down vote
All right, I realized that representing $arctan(x)$ through the integral of an oscillating function is not a good idea. Better to represent both $arctan(x)$ and $textSi(x)$ as integrals of monotonic and easily-comparable functions. So here it is a polished version of the previous answer. We may safely assume $x>1$ since power series easily prove the statement for $xin[0,1]$. By the Laplace transform and the Cauchy-Schwarz inequality
$$ textSi(x)=fracpi2-int_0^+inftyfraccos(x)+ssin(x)(1+s^2)e^sx,dsgeq fracpi2-int_0^+inftyfracdse^sxsqrt1+s^2. tag1$$
By the very definition of $arctan$ we have $arctan(x)=left(int_0^+inftyfrac11+s^2-frac11+(s+x)^2right),dx$, hence through $arctan x=fracpi2-arctanfrac1x$ we get the following integral representation:
$$ arctan(x) = fracpi2-int_0^+inftyfrac1+2sx(1+s^2)(1+2sx+x^2+s^2 x^2),ds. tag2$$
For the sake of brevity, let us denote as $S(x,s)$ and $T(x,s)$ the integrand functions appearing in the RHSs of $(1)$ and $(2)$. If we manage to prove $S(x,s)leq T(x,s)$ for any $x>1$ and any $s>0$ we are done. But the Padé approximants for the exponential function reveal that this is a pretty loose inequality, so we are good to go:
$$ forall x>0,qquad textSi(x)>arctan(x).tag3 $$
A strange-looking consequence of $(1)$ and the AM-QM inequality is also
$$ textSi(x) > fracpi2-sqrt2,e^x,Gamma(0,x).tag4$$
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
 |Â
show 8 more comments
up vote
6
down vote
up vote
6
down vote
All right, I realized that representing $arctan(x)$ through the integral of an oscillating function is not a good idea. Better to represent both $arctan(x)$ and $textSi(x)$ as integrals of monotonic and easily-comparable functions. So here it is a polished version of the previous answer. We may safely assume $x>1$ since power series easily prove the statement for $xin[0,1]$. By the Laplace transform and the Cauchy-Schwarz inequality
$$ textSi(x)=fracpi2-int_0^+inftyfraccos(x)+ssin(x)(1+s^2)e^sx,dsgeq fracpi2-int_0^+inftyfracdse^sxsqrt1+s^2. tag1$$
By the very definition of $arctan$ we have $arctan(x)=left(int_0^+inftyfrac11+s^2-frac11+(s+x)^2right),dx$, hence through $arctan x=fracpi2-arctanfrac1x$ we get the following integral representation:
$$ arctan(x) = fracpi2-int_0^+inftyfrac1+2sx(1+s^2)(1+2sx+x^2+s^2 x^2),ds. tag2$$
For the sake of brevity, let us denote as $S(x,s)$ and $T(x,s)$ the integrand functions appearing in the RHSs of $(1)$ and $(2)$. If we manage to prove $S(x,s)leq T(x,s)$ for any $x>1$ and any $s>0$ we are done. But the Padé approximants for the exponential function reveal that this is a pretty loose inequality, so we are good to go:
$$ forall x>0,qquad textSi(x)>arctan(x).tag3 $$
A strange-looking consequence of $(1)$ and the AM-QM inequality is also
$$ textSi(x) > fracpi2-sqrt2,e^x,Gamma(0,x).tag4$$
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
All right, I realized that representing $arctan(x)$ through the integral of an oscillating function is not a good idea. Better to represent both $arctan(x)$ and $textSi(x)$ as integrals of monotonic and easily-comparable functions. So here it is a polished version of the previous answer. We may safely assume $x>1$ since power series easily prove the statement for $xin[0,1]$. By the Laplace transform and the Cauchy-Schwarz inequality
$$ textSi(x)=fracpi2-int_0^+inftyfraccos(x)+ssin(x)(1+s^2)e^sx,dsgeq fracpi2-int_0^+inftyfracdse^sxsqrt1+s^2. tag1$$
By the very definition of $arctan$ we have $arctan(x)=left(int_0^+inftyfrac11+s^2-frac11+(s+x)^2right),dx$, hence through $arctan x=fracpi2-arctanfrac1x$ we get the following integral representation:
$$ arctan(x) = fracpi2-int_0^+inftyfrac1+2sx(1+s^2)(1+2sx+x^2+s^2 x^2),ds. tag2$$
For the sake of brevity, let us denote as $S(x,s)$ and $T(x,s)$ the integrand functions appearing in the RHSs of $(1)$ and $(2)$. If we manage to prove $S(x,s)leq T(x,s)$ for any $x>1$ and any $s>0$ we are done. But the Padé approximants for the exponential function reveal that this is a pretty loose inequality, so we are good to go:
$$ forall x>0,qquad textSi(x)>arctan(x).tag3 $$
A strange-looking consequence of $(1)$ and the AM-QM inequality is also
$$ textSi(x) > fracpi2-sqrt2,e^x,Gamma(0,x).tag4$$
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
edited Aug 11 at 5:20
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
answered Aug 11 at 4:12
Jack D'Aurizio♦
271k31266632
271k31266632
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
 |Â
show 8 more comments
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^-1left(frac1sarctanfrac1sright)(x)=frac12pi iint_-iinfty^iinftye^sxfrac1sarctanfrac1sds$?
– Ramanasa
Aug 11 at 5:09
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
@Ramanasa: $$ textSi(x) = int_0^+inftyfrac1tcdotsin(t)mathbb1_(0,x)(t),dt $$ if we apply $mathcalL^-1$ to $frac1t$ and $mathcalL$ to $sin(t)mathbb1_(0,x)(t)$, then recall that $int_0^+inftyfracsin tt,dt=fracpi2$, we reach the integral representation at the beginning of $(1)$.
– Jack D'Aurizio♦
Aug 11 at 5:14
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/…
– Jack D'Aurizio♦
Aug 11 at 5:15
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance.
– Jack D'Aurizio♦
Aug 11 at 5:19
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ operatornameSi(x)=int_0^infty frac1tcdotsin t cdot1_(0,infty)dt $$ what exactly mean to apply $L^-1$ to $frac1t$ and $L$ to $sin t cdot 1_(0,infty)$? I do not know what property of laplace transform you used here.
– Ramanasa
Aug 11 at 5:46
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$$textSi'(x) = fracsin xx = fracsum_k (-1)^kx^2k+1/(2k+1)!x = sum_k (-1)^kx^2k/(2k+1)!$$
$$arctan'(x) = frac11+x^2 = frac11-(-x^2) = sum_k (-x^2)^k = sum_k (-1)^kx^2k$$
$$textSi(x) = sum_k frac(-1)^kx^2k+1/(2k+1)(2k+1)!$$
$$arctan(x) = sum_k frac(-1)^kx^2k+1/(2k+1)1$$
$$frac1(2k+1)! geq frac11$$
This would give the opposite of your inequality... but we must take into account the $(-1)^k$.
answered Aug 11 at 3:42
mr_e_man
854420
3
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
add a comment |Â
up vote
0
down vote
$$textSi'(x) = fracsin xx = fracsum_k (-1)^kx^2k+1/(2k+1)!x = sum_k (-1)^kx^2k/(2k+1)!$$
$$arctan'(x) = frac11+x^2 = frac11-(-x^2) = sum_k (-x^2)^k = sum_k (-1)^kx^2k$$
$$textSi(x) = sum_k frac(-1)^kx^2k+1/(2k+1)(2k+1)!$$
$$arctan(x) = sum_k frac(-1)^kx^2k+1/(2k+1)1$$
$$frac1(2k+1)! geq frac11$$
This would give the opposite of your inequality... but we must take into account the $(-1)^k$.
answered Aug 11 at 3:42
mr_e_man
854420
3
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$textSi'(x) = fracsin xx = fracsum_k (-1)^kx^2k+1/(2k+1)!x = sum_k (-1)^kx^2k/(2k+1)!$$
$$arctan'(x) = frac11+x^2 = frac11-(-x^2) = sum_k (-x^2)^k = sum_k (-1)^kx^2k$$
$$textSi(x) = sum_k frac(-1)^kx^2k+1/(2k+1)(2k+1)!$$
$$arctan(x) = sum_k frac(-1)^kx^2k+1/(2k+1)1$$
$$frac1(2k+1)! geq frac11$$
This would give the opposite of your inequality... but we must take into account the $(-1)^k$.
answered Aug 11 at 3:42
mr_e_man
854420
$$textSi'(x) = fracsin xx = fracsum_k (-1)^kx^2k+1/(2k+1)!x = sum_k (-1)^kx^2k/(2k+1)!$$
$$arctan'(x) = frac11+x^2 = frac11-(-x^2) = sum_k (-x^2)^k = sum_k (-1)^kx^2k$$
$$textSi(x) = sum_k frac(-1)^kx^2k+1/(2k+1)(2k+1)!$$
$$arctan(x) = sum_k frac(-1)^kx^2k+1/(2k+1)1$$
$$frac1(2k+1)! geq frac11$$
This would give the opposite of your inequality... but we must take into account the $(-1)^k$.
answered Aug 11 at 3:42
mr_e_man
854420
edited Aug 11 at 3:50
answered Aug 11 at 3:42
mr_e_man
854420
answered Aug 11 at 3:42
mr_e_man
854420
answered Aug 11 at 3:42
mr_e_man
854420
854420
3
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
add a comment |Â
3
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
3
3
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
While $textSi(x)$ is an entire function, the radius of convergence of the Maclaurin series of $arctan(x)$ is just $1$, so these series do not allow to compare $textSi(x)$ and $arctan(x)$ outside $|x|leq 1$.
– Jack D'Aurizio♦
Aug 11 at 3:50
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
That is true. This answer is incomplete anyway.
– mr_e_man
Aug 11 at 3:52
add a comment |Â
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I don't know if this helps but maybe try to express $arctan(x)$ as the integral from $0$ to $x$ of its derivative $frac1t^2+1$ and then form one integral.
– zzuussee
Aug 11 at 3:26
I tried to but could not proceed more. It seems there needs some more manipulations.
– Ramanasa
Aug 11 at 3:34
I'm currently writing a sketch of some thoughts.
– zzuussee
Aug 11 at 3:34
If you can deal with power series, then write both $operatornameSi(x)$ and $arctan (x)$ as power series and note their difference when $x>0.$
– Allawonder
Aug 11 at 3:43
2
@zzuussee: but a faulty one, since $arctan(x)$ is not an entire function. On the other hand an interesting relation between $arctan$ and $textSi$ is $$ mathcalLleft(textSi(x)right)(s) = frac1sarctanfrac1s.$$
– Jack D'Aurizio♦
Aug 11 at 3:52